Determine the pH of 8.50×10−2 M KCN?
- hcbiochemLv 72 months ago
CN- acts as a base by the equilibrium:
CN- + H2O <--> HCN + OH-
Kb = [HCN][OH-]/[HCN]
Kb can be calculated from Ka of HCN as:Kb = 1.00X10^-14 / Ka
Kb = 1.00X10^-14 / 6.2X10^-10 = 1.6X10^-5
In the solution, let [HCN] = [OH-] = x, and [CN-] = 8.50X10^-2 - x.
As a first approximation, assume that x will be small compared to 8.50X10^-2. Then,
Kb = 1.6X10^-5 = x^2/0.0850
x = 1.2X10^-3 = [OH-]
pOH = 2.93
pH = 14.00 - pOH = 11.07